# Double orthogonal complement of any closed subspace is it self [duplicate]

Let $H$ be a pre-Hilbert space such that any closed sub space $M \subset H$ has the property $M^{\bot \bot}=M$. Prove that $H$ is a Hilbert space (ie, prove that $H$ is complete)

My attempt:

As $H^*$ ($L(H,\mathbb{K})$) is complete then I want to prove that $H \cong H^*$. But I don't know how to construct an isometry between them.

I will appreciate highly who can give me some ideas

That space seems to be the same space here: A counterexample to theorem about orthogonal projection and it doens´t seems to have the property $$M = M^{\perp \perp}$$ for all closed subspaces $$M$$.

I found the same exercise in the book "Functional Analysis" of George Bachman and Lawrence Narici

It says: If X is an inner product space and $$M=M^{\perp \perp}$$ for every closed subspace M of X, show that X is a Hilbert space. [Hint Use the mapping T mentioned in Eq. (12.23)]

That mapping is:

$$T:X \rightarrow \overset{\sim}{X}$$ such that

$$y \rightarrow Ty = f$$

where the bounded functional $$f$$ is, for any given $$x_{0} \in X$$, given by:

$$(Ty)(x)=f(x)=(x,y)$$

Edit: I think I was able to prove it, I will explain it now.

Let's be $$T:X \to \overset{\sim}{X}$$ such that $$T(y)=f_{y}$$ where $$f_{y}(x)=(x,y) \forall x \in X$$. That is, $$T(y)$$ is the Riesz representative of $$y$$.

And $$\overline{T} : \overline{X} \to \overset{\sim}{\overline{X}}$$ to the same function but with domain $$\overline{X}$$ (the completion of X).

The conjugate space a space and it's completion are always isomorphic, so we will identify both conjugate spaces and act as if $$\overset{\sim}{ X } = \overset{\sim}{\overline{X}}$$.

Observe that the function $$\overline{T}$$ is an isometry, so it's injective. Then $$T = \overline{T}_{|X}$$ it's also injective. I will make a strong use of this property, that's why I am pointing it out.

We will prove it by contradiction. Assume that there exists $$x \in \overline{X} - X$$.

As every vector space contains the element $$0$$, $$x_{0}$$ cannot be $$0$$.

Let $$f=\overline{T} (x_{0})$$, as $$x_{0} \neq 0$$ and $$\overline{T}$$ is injective, we know that $$f \neq 0$$.

$$N_{f}= \{ z \in X / f(z)=0 \} = f^{-1} ( \{ 0 \} )$$ is a closed subspace of $$X$$, as it's the inverse image of $$0$$ under $$f$$ and $$f$$ is continuous.

As $$f \neq 0$$, $$f$$ doesn't vanish in all $$X$$, that is $$\begin{equation} \label{Hola} N_{f} \neq X \end{equation}$$

We will use that inequality to arrive a contradiction. If $$N_{f}^{\perp} = \{ 0 \}$$ then $$N_{f}=N_{f}^{\perp \perp} = (N_{f}^{\perp})^{\perp} = \{ 0 \}^{\perp} = X$$ Where the first equality is given by hyphotesis. So if we can prove that $$N_{f}^{\perp} = \{ 0 \}$$ we will have reached a contradiction.

By definition $$N_{f}^{\perp} = \{ x \in X / (y,x) = 0 \: \forall \: y \in N_{f} \}$$

Let $$x$$ be an arbitrary element of $$N_{f}$$, lets show that $$x = 0$$.

$$x \in N_{f}^{\perp}$$ is the same than saying $$(y,x) = 0$$ for all $$y \in N_{f}$$ which implies $$T(x)(y)=0$$ for all $$y \in N_{f}$$. Note that $$f$$ is the only nonzero functional which vanish in all $$N_{f}$$ modulo a nonzero scalar, this implies either $$T(x) = \lambda f =\lambda T(x_{0})$$ for a nonzero $$\lambda$$ or $$T(x) = 0$$ which implies $$x=0$$. Let's discard the first option. In this case $$T( \lambda^{-1} x ) = T(x_{0})$$, and being $$T$$ injective this implies $$\lambda^{-1} x = x_{0}$$.

Note that $$span( x_{0} ) \cap X = \{ 0 \}$$ (otherwise we would have $$\lambda x_{0} \in X$$ with $$\lambda \neq 0$$ so $$x_{0} = \lambda^{-1} \lambda x_{0} \in X$$ which contradicts the initial assumption that $$x_{0} \notin X$$).

So we have two facts. $$0 \neq x_{0} = \lambda^{-1} x \in span (x_{0} ) \cap X$$ and $$\{ 0 \} = span (x_{0} ) \cap X$$, which is a contradiction.

As this contradiction arised from assuming that $$x \neq 0$$, it must be have $$x=0$$. As $$x$$ was an arbitrary element of $$N_{f}^{\perp}$$ we just have proved $$N_{f}^{\perp} = \{ 0 \}$$.

We knew that if $$N_{f}^{\perp} = \{ 0 \}$$ we would reach a contradiction, thus we arrived a contradiction in either case. This contradiction arised supposing that $$X \neq \overline{X}$$, so it must be$$X=\overline{X}$$. Therefore $$X$$ is a Hilbert Space, as is a complete inner product space.

Comment after the edit: I wrote this years ago in Spanish, it took me several tries to write a understandable proof. Then I completely butchered it while trying to translate it to English with my poor English skills. I am really sorry for posting such a illegible mess. I don't remember the proof nor I have the sheet in which I have written it, so I had to re-read it and try to make sense of it. The structure it's quite convoluted, so a mistake may have slipped of my sight. I hope it's correct now, I will read it again this week.

• It doesn't look like the argument in the proof is sound. Besides the fact that it is not clear what $x$ and $x_0$ are, it is also not clear what contradicts what. Jan 17, 2022 at 21:39
• Hey @MartinArgerami thanks for bringing this to my attention. You're right, nothing was clear. I am not sure if it's correct, but I hope it's at least readable now. Jan 17, 2022 at 23:49
• Thanks. I understand the argument, but I still think it is written in a very unclear way. Compare to Jochen's answer to the other question. Jan 18, 2022 at 13:50
• Oh, thank you. I will be sincere, On the 2015 I spent 3 months attacking this problem without any luck, and when I was finally succesful I wrote a convoluted proof on Spanish and made it more convoluted (and incorrect) on English. Now, I didn't think the problem again (as it intimidates me), I merely tried to correct my proof, which of course didn't de-convolute it. But I am happy to know that there is an easier way, I will look at it. Jan 18, 2022 at 13:53

This statement appears to be false:

Let $H=\ell^2(\mathbb R)$ be the Hilbert space with the usual orthonormal basis sequence $(e_n)_{n \in \mathbb N}$. Let $L$ denote the linear span of $e_n$.

Then $L$ is an (incomplete) inner product space: $$s_n =\sum_{k=0}^n {1\over 2^k}e_k$$

converges to the sequence $s = (1,{1\over 2}, {1\over 4},\dots)$ which is not in $L$ but $s_n \in L$.

Since $e_n$ are orthonormal it is easy to see that if $M$ is any subspace of $L$ then $M^{\bot \bot}=M$.

• It's easy to say "easy to see". Do you have a proof? I don't think that your statement is correct. Dec 29, 2016 at 23:00